010. Regular Expression Matching.cpp

// Solution 1. Recursion
class Solution {
public:
    bool isMatch(string s, string p) {
        if(p.empty()) return s.empty();
        if(s.empty()) return p.empty() || (p[1] == '*' ? isMatch(s, p.substr(2)) : false);
        if(p[0] != '.' && s[0] != p[0]) return p[1] == '*' ? isMatch(s, p.substr(2)) : false;
        if(p[1] == '*') return isMatch(s.substr(1), p) || isMatch(s, p.substr(2));
        return isMatch(s.substr(1), p.substr(1));
    }
};

// Solution 2. DP
class Solution {
public:
    bool isMatch(string s, string p) {
        /**
         * f[i][j]: if s[0..i-1] matches p[0..j-1]
         * if p[j - 1] != '*'
         *      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
         * if p[j - 1] == '*', denote p[j - 2] with x
         *      f[i][j] is true iff any of the following is true
         *      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
         *      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
         * '.' matches any single character
         */
        int m = s.size(), n = p.size();
        vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));
        
        f[0][0] = true;
        for (int i = 1; i <= m; i++)
            f[i][0] = false;
        // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty
        for (int j = 1; j <= n; j++)
            f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2];
        
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (p[j - 1] != '*')
                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
                else
                    // p[0] cannot be '*' so no need to check "j > 1" here
                    f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];
        
        return f[m][n];
    }
};